Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{-5n^3 - 45n^2 - 100n}{-n^3 + 4n^2 - 4n} \div \dfrac{3n + 15}{-n + 2} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-5n^3 - 45n^2 - 100n}{-n^3 + 4n^2 - 4n} \times \dfrac{-n + 2}{3n + 15} $ First factor out any common factors. $q = \dfrac{-5n(n^2 + 9n + 20)}{-n(n^2 - 4n + 4)} \times \dfrac{-(n - 2)}{3(n + 5)} $ Then factor the quadratic expressions. $q = \dfrac {-5n(n + 5)(n + 4)} {-n(n - 2)(n - 2)} \times \dfrac {-(n - 2)} {3(n + 5)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { -5n(n + 5)(n + 4) \times -(n - 2)} { -n(n - 2)(n - 2) \times 3(n + 5)} $ $q = \dfrac {5n(n + 5)(n + 4)(n - 2)} {-3n(n - 2)(n - 2)(n + 5)} $ Notice that $(n - 2)$ and $(n + 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {5n(n + 5)(n + 4)\cancel{(n - 2)}} {-3n\cancel{(n - 2)}(n - 2)(n + 5)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $q = \dfrac {5n\cancel{(n + 5)}(n + 4)\cancel{(n - 2)}} {-3n\cancel{(n - 2)}(n - 2)\cancel{(n + 5)}} $ We are dividing by $n + 5$ , so $n + 5 \neq 0$ Therefore, $n \neq -5$ $q = \dfrac {5n(n + 4)} {-3n(n - 2)} $ $ q = \dfrac{-5(n + 4)}{3(n - 2)}; n \neq 2; n \neq -5 $